This Exam P sample reference tests Exponential Distribution. Each exponential operating time has variance 10 squared. Independence makes the total variance 100+100=200, choice E.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AChoice A reports the mean of one generator rather than a variance.
BChoice B reports the mean of the combined operating time.
CChoice C includes only half of the required total variance.
DChoice D includes the variance of only one exponential operating time.
Original practice · fully worked
Original variant: variance of a two-stage service time
A data request passes through two independent service stages. Their times are exponential with means 4 seconds and 7 seconds. Find the variance of total service time.
A 11 seconds squared
B 33 seconds squared
C 49 seconds squared
D 65 seconds squared
E 121 seconds squared
Variant answer in brief
Exponential variance equals mean squared. Adding the independent stage variances gives 16+49=65.
Setup
Setup
The stage variances are the squares of the exponential means, giving 16 and 49 seconds squared.
Var(X)=42=16
Var(Y)=72=49
Model
Model
Independence removes the covariance term, so the total service-time variance is the sum of the two stage variances.
Var(X+Y)=16+49
Compute
Compute
The resulting variance is 65 seconds squared.
Var(X+Y)=65
Answer
Answer
The total service time has variance 65 seconds squared, selecting choice D.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.